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Chapter 4: Problem 93

A penguin slides at a constant velocity of \(1.4 \mathrm{m} / \mathrm{s}\) down an icy incline. The incline slopes above the horizontal at an angle of \(6.9^{\circ} .\) At the bottom of the incline, the penguin slides onto a horizontal patch of ice. The coefficient of kinetic friction between the penguin and the ice is the same for the incline as for the horizontal patch. How much time is required for the penguin to slide to a halt after entering the horizontal patch of ice?

Short Answer

Expert verified
The penguin will take approximately 1.18 seconds to come to a halt.

Step by step solution

01

Analyze the Problem

We have a penguin sliding at a constant velocity down an incline and then onto a horizontal surface where it eventually comes to a stop. The angle of the incline and the coefficient of kinetic friction will help us determine the force of friction acting on the penguin on the horizontal surface.
02

Determine the Acceleration on the Horizontal Patch

Since the penguin will come to a stop, its initial velocity is 1.4 m/s, and the finally it will be 0 m/s. Using the formula for acceleration due to friction, we have, on the horizontal surface:\[ f_k = \mu_k \cdot m \cdot g = m \cdot a \] Solving for acceleration (\(a\)), we get:\[ a = \mu_k \cdot g \]However, we need to find the coefficient of kinetic friction \(\mu_k\) from the inclined plane analysis first.
03

Calculate the Coefficient of Kinetic Friction

On the inclined plane, since the penguin slides at constant velocity, the net force is zero. Thus, the gravitational component down the slope is balanced by friction:\[ m \cdot g \cdot \sin(6.9^{\circ}) = \mu_k \cdot m \cdot g \cdot \cos(6.9^{\circ}) \]The masses cancel out, and we can solve for \(\mu_k\):\[ \mu_k = \tan(6.9^{\circ}) \approx 0.121 \]
04

Solve for Time of Sliding

With the coefficient of kinetic friction, we substitute into the equation for acceleration on the horizontal surface:\[ a = \mu_k \cdot g = 0.121 \cdot 9.8 \approx 1.1868 \mathrm{m/s^2} \]Now using the kinematic equation:\[ v_f = v_i + at \]Given \(v_f = 0\), \(v_i = 1.4 m/s\), and solving for \(t\):\[ 0 = 1.4 - 1.1868 \cdot t \]\[ t = \frac{1.4}{1.1868} \approx 1.18 \text{ seconds} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Constant Velocity
When we talk about constant velocity, we're discussing a situation where an object moves at the same speed and in the same direction continuously. For instance, the penguin in our exercise slides down the incline at a constant velocity of 1.4 m/s. This means that the forces acting on the penguin are perfectly balanced.

Understanding this balance is crucial, as it tells us that the gravitational pull down the slope and the frictional force opposing it are equal.
  • This equilibrium results in no net acceleration.
  • The penguin's speed doesn't change as long as it stays on the incline.
Keeping things simple, constant velocity implies no change in speed unless an external force acts on the object.
Incline
The incline is a sloped surface that creates a unique dynamic for motion due to its angle relative to horizontal ground. In the exercise, the incline angle is crucial at 6.9 degrees, influencing the forces acting on the penguin.

There are two main components to consider:
  • The component of gravitational force pulling the penguin down the slope.
  • The normal force perpendicular to the surface, which affects friction.
The angle impacts not just the motion but also how a coefficient of kinetic friction is perceived. Down the slope, gravitational force can be split into parallel and perpendicular components. The parallel component (along the slope), given by \( m \cdot g \cdot \sin(6.9^{\circ}) \), drives the sliding motion. The perpendicular one affects the frictional force.
Acceleration
Acceleration is about changes in velocity. In our problem, it becomes particularly interesting once the penguin reaches the horizontal plane. While on the incline, the penguin accelerates due to gravity's component down the slope. On the flat surface, however, it decelerates due to friction.

To find the acceleration (which is negative in this case, as it's a deceleration), we use the formula:\[ a = \mu_k \cdot g \]This equation shows us how the friction coefficient and gravitational force combine to slow the penguin.
  • When the penguin hits the horizontal ice, friction is the only force acting to change its velocity.
  • Deceleration here leads to a reduction in speed till the penguin halts.
This understanding of acceleration helps in predicting how quickly an object will come to rest given initial motion conditions.
Kinetic Friction Coefficient
The kinetic friction coefficient, denoted as \( \mu_k \), is a measure of the friction between moving surfaces. For our scenario, it's crucial because it affects how quickly the penguin stops sliding after reaching the bottom of the incline.

To find this coefficient, we relied on the constant velocity condition on the incline. The equation:\[ \mu_k = \tan(6.9^{\circ}) \approx 0.121 \]shows how the incline's angle determines the coefficient.
  • This value of \( \mu_k \) applies to both the incline and horizontal surface.
  • It indicates the relationship between gravitational pull and frictional force, ensuring balance in motion and transition to a halt.
Knowing \( \mu_k \) is key to solving questions around how much time or distance is needed for motion changes, such as coming to a stop after a slide on different surfaces.

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Most popular questions from this chapter

During a storm, a tree limb breaks off and comes to rest across a barbed wire fence at a point that is not in the middle between two fence posts. The limb exerts a downward force of 151 N on the wire. The left section of the wire makes an angle of \(14.0^{\circ}\) relative to the horizontal and sustains a tension of \(447 \mathrm{N}\). Find the magnitude and direction of the tension that the right section of the wire sustains.

Consult Multiple-Concept Example 10 for insight into solving this type of problem. A box is sliding up an incline that makes an angle of \(15.0^{\circ}\) with respect to the horizontal. The coefficient of kinetic friction between the box and the surface of the incline is \(0.180 .\) The initial speed of the box at the bottom of the incline is \(1.50 \mathrm{m} / \mathrm{s}\). How far does the box travel along the incline before coming to rest?

On earth, two parts of a space probe weigh 11 000 N and 3400 N. These parts are separated by a center-to-center distance of 12 m and may be treated as uniform spherical objects. Find the magnitude of the gravitational force that each part exerts on the other out in space, far from any other objects.

A 15-g bullet is fired from a rifle. It takes \(2.50 \times 10^{-3}\) s for the bullet to travel the length of the barrel, and it exits the barrel with a speed of \(715 \mathrm{m} / \mathrm{s} .\) Assuming that the acceleration of the bullet is constant, find the average net force exerted on the bullet.

Scientists are experimenting with a kind of gun that may eventually be used to fi re payloads directly into orbit. In one test, this gun accelerates a 5.0-kg projectile from rest to a speed of \(4.0 \times 10^{3} \mathrm{m} / \mathrm{s} .\) The net force accelerating the projectile is \(4.9 \times 10^{5} \mathrm{N} .\) How much time is required for the projectile to come up to speed?
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